For the first-order reaction half-life is 14 s. The time required for the initial concentration to reduce to 1/8th of its value is:
Answer & explanation
Correct answer: option 3
The correct answer is option 3. 42 s.
Given,
Half-life, \(t_{1/2} = 14 s\)
We know,
\(t_{1/2} = \frac{0.693}{k}\)
or, \(k = \frac{0.693}{t_{1/2}}\)
\(⇒ k =\frac{0.693}{14}\)
\(∴ k = 0.0495\)
Let,'a' be the initial concentration, then according to the question,
final concentration \(= \frac{a}{8}\)
\(∴ k = \frac{2.303}{t}log \frac{a}{a/8}\)
\(⇒ t = \frac{2.303}{k}log \frac{a}{a/8}\)
\(⇒ t = \frac{2.303}{0.0495}log 8\)
\(⇒ t = \frac{2.303}{0.0495}× 0.903\)
\(⇒ t = 42.01 s\)
\(∴ t ≈ 42 s\)